You fire a 1kg teflon ball from a fur-lined cannon near the surface of the earth
ID: 2273701 • Letter: Y
Question
You fire a 1kg teflon ball from a fur-lined cannon near the surface of the earth The ball emerges from the cannon with an excess surface charge of -5mC, at a muzzle velocity of 120 m/s, and 30 degrees above the horizontal. The cannon is fired into a region of a uniform 100 N/C electric field pointing 45 degrees below the horizontal as show below. How much higher (or lower) and further (or shorter) does the teflon ball travel due to the electric field? Ignore frictional effects of the air, assume the teflon ball does not lose any excess charge, and ignore the height of the cannon.
You fire a 1kg teflon ball from a fur-lined cannon near the surface of the earth The ball emerges from the cannon with an excess surface charge of -5mC, at a muzzle velocity of 120 m/s, and 30 degrees above the horizontal. The cannon is fired into a region of a uniform 100 N/C electric field pointing 45 degrees below the horizontal as show below. How much higher (or lower) and further (or shorter) does the teflon ball travel due to the electric field? Ignore frictional effects of the air, assume the teflon ball does not lose any excess charge, and ignore the height of the cannon.Explanation / Answer
F = qE
F=5*10^-3 * 100 =0.5 N
acceleration = 0.5N/1kg = 0.5m/s^2
when this acceleration is resolved into components..we get a =- 0.5/sqrt2 i - 0.5/sqrt2 j
it travels for less time than it is supposed to be..
total time now = 2 * 120*sin 30/(9.8+ 0.5/sqrt2) = 11.818 sec
time if der was no field = 2 * 120*sin 30/(9.8) =12.244 sec
distance travelld if der was no field = 120*cos 30 * 11.818 = 1228.1 m
actual distance = (120*cos 30- 0.5/sqrt2) * 12.244 sec = 1268 m
xtra distance = 39.9 m
xtra time = 0.426 sec
xtra height = (120*sin30)^2/(2*9.8) - (120*sin30)^2/(2*(9.8+0.5/sqrt2))
= 171.36m
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